By Stanley Burris

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*Mathematical Reviews*of the yankee Mathematical Society.

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45 §5. Congruences and Quotient Algebras By symmetry c, a ∈ θ1 ; hence c, a ∈ θ2 , and then by transitivity c, b ∈ θ2 . Thus c, b ∈ θ2 ∩ θ3 , so from aθ1 c(θ2 ∩ θ3 )b follows a, b ∈ θ1 ◦ (θ2 ∩ θ3 ); hence a, b ∈ θ1 ∨ (θ2 ∩ θ3 ). ✷ We would like to note that in 1953 J´onsson improved on Birkhoﬀ’s result above by showing that one could derive the so-called Arguesian identity for lattices from congruencepermutability. In §12 we will concern ourselves again with congruence-distributivity and permutability.

Then the composition β ◦ α is a homomorphism from A to C. Proof. For f an n-ary function symbol and a1 , . . , an ∈ A, we have (β ◦ α)f A(a1 , . . , an ) = β(αf A(a1 , . . , an )) = βf B (αa1 , . . , αan ) = f C (β(αa1 ), . . , β(αan )) = f C ((β ◦ α)a1 , . . , (β ◦ α)an ). ✷ The next result says that homomorphisms commute with subuniverse closure operators. 6. If α : A → B is a homomorphism and X is a subset of A then α Sg(X) = Sg(αX). Proof. From the deﬁnition of E (see §3) and the fact that α is a homomorphism we have αE(Y ) = E(αY ) for all Y ⊆ A.

Let K be a class of algebras and let K1 be a proper subclass of K. ) Two basic questions arise in the quest for structure theorems. (1) Is every member of K isomorphic to some member of K1 ? (2) Is every member of K embeddable in some member of K1 ? For example, every Boolean algebra is isomorphic to a ﬁeld of sets (see IV§1), every group is isomorphic to a group of permutations, a ﬁnite Abelian group is isomorphic to a direct product of cyclic groups, and a ﬁnite distributive lattice can be embedded in a power of the two-element distributive lattice.